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Stokes equation in a toy CD hovercraft
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2011 Eur. J. Phys. 32 89
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IOP PUBLISHING
EUROPEAN JOURNAL OF PHYSICS
Eur. J. Phys. 32 (2011) 89–99
doi:10.1088/0143-0807/32/1/008
Stokes equation in a toy CD hovercraft
Charles de Izarra and Grégoire de Izarra
Groupe de Recherche sur l’Énergetique des Milieux Ionisés, UMR6606 Université
d’Orléans—CNRS, Faculté des Sciences, Site de Bourges, rue Gaston Berger, BP 4043,
18028 Bourges Cedex, France
E-mail: Charles.De_Izarra@univ-orleans.fr
Received 25 July 2010, in final form 20 September 2010
Published 17 November 2010
Online at stacks.iop.org/EJP/32/89
Abstract
This paper deals with the study of a toy CD hovercraft used in the fluid
mechanics course for undergraduate students to illustrate the lubrication theory
described by the Stokes equation. An experimental characterization of the toy
hovercraft (measurements of the air flow value, of the pressure in the balloon
and of the thickness of the air film under the hovercraft) allows us to evaluate
a reduced Reynolds number R ∗ . Since R ∗ < 1, it is possible to simplify the
Navier–Stokes equation that is reduced to the Stokes equation, on the basis
of the lubrication theory. The pressure gradient in the air flow is calculated,
allowing us to establish the lifting force applied on the toy hovercraft. In
addition, these results are applied to a larger scale hovercraft.
(Some figures in this article are in colour only in the electronic version)
1. Introduction
A lesson on fluid mechanics for undergraduate students generally begins with fluid statics,
followed by the study of ideal fluid dynamics described by the Euler equation. Introduction
of the fluid viscosity leads to the Navier–Stokes equation that can be analytically solved only
in a very reduced number of cases [1]. When the inertial force is negligible in front of the
viscous force, the Navier–Stokes equation describes fluid dynamics at a low Reynolds number,
which is of great importance in many fields of sciences, from lubrication theory to the study
of Paramecia motion [2]. A possible course demonstration that illustrates the fluid effects at
low Reynolds number is obtained by dropping a sheet of paper on a table: the thin air layer
between the table and the paper sheet produces an important pressure variation and a lifting
force; the paper sheet glides on the table. In this paper, we propose a small science toy easily
built that may be used for a demonstration course or for a fluid mechanics exercise.
2. Experiment: experimental results
2.1. Building a toy CD hovercraft
The construction of a toy CD hovercraft needs a CD, a medium-sized balloon and a sports
bottle lid that should be opened by pulling the nozzle and closed by performing the same
c 2011 IOP Publishing Ltd Printed in the UK & the USA
0143-0807/11/010089+11$33.00 89
90
C de Izarra and G de Izarra
Figure 1. Construction of a toy CD hovercraft with a CD, a sports bottle lid and a balloon.
action (figure 1). The bottle lid is glued over the hole of the CD, and the balloon is attached to
the opening of the lid. To operate the hovercraft, it is necessary to open the sports bottle lid, to
blow air into the balloon through the CD hole, and to close the lid so that the air cannot escape.
The CD is placed on a smooth surface, the lid is carefully opened with a small aperture and
the hovercraft glides slowly on the surface.
2.2. Assumptions and measurements
Different measurements were performed on the toy CD hovercraft for which the balloon was
chosen spherical. The total mass of the system filled with air is m = 21.45 g. Using a pressure
sensor, we have checked that the overpressure PB in the balloon was nearly constant during
the time when the balloon empties: PB = (1000 ± 100) Pa.
The volume V of air contained in the balloon at pressure PB +PA , PA being the atmospheric
pressure, was determined by measuring the diameter of the spherical balloon. The time
needed to completely empty a balloon through the small aperture of the lid is about 15 s;
consequently, the process can be considered isothermal. The volume V of air contained in the
balloon evaluated for the atmospheric pressure PA is given by V = (PA + PB )V /PA . Since
PB = 0.01PA , it is clear that V is very close to V , and we can take V = V .
Between the cylinder of radius R0 = 6 mm located in the central part of the CD and the
cylinder of radius R1 = 60 mm, we can consider an air flow with a cylindrical symmetry
located in a thin film of height h (figures 2 and 3). The pressure value P0 in the central part of
the system (cylinder of radius R0) is lower than the pressure inside the balloon; the value of
P0 depends on the flow value, and will be calculated later.
The thickness h of the air film between the CD surface and the table was measured
using two methods. The first method uses sheets of adhesive papers whose thicknesses were
Stokes equation in a toy CD hovercraft
91
z
R1
R0
y
h
x
Figure 2. The toy CD hovercraft on a smooth surface.
PB + PA
P0
h
PA
Figure 3. Air flow between the CD and the smooth surface (lateral view along a diameter).
determined using a micrometer screw gauge, successively placed on a sheet of paper put on the
table. When the toy CD hovercraft is stopped by the paper, we can consider that the thickness
of the paper is higher than h. This simple method indicates an h value located between
0.4 and 0.6 mm. A second method uses a non-contact optical displacement measuring system
(OPTONCDT LD 1605-20 from MICRO-EPSILON); we obtain h = (0.55 ± 0.05) mm.
To measure the volumetric air flow Q, we have applied the following procedure that needs
a chronometer and two operators. First, the spherical balloon is inflated, and its circumference
is measured using a flexible tape measure. The circumference of the balloon is used to calculate
the volume of air V it contains. The first operator places the hovercraft on a flat surface and
carefully opens the lid. Simultaneously, the second operator starts a timer and stops when the
balloon is empty to measure the time τ . The volumetric air flow is given by Q = V /τ . We
obtain the value 0.3 × 10−3 m3 s−1 . Considering the flow incompressible (justified because
the Mach number is lower than unity), we have Q = v0 S0 where v 0 is a mean air velocity
value at the radius R0 and S0 = 2π hR0 is the lateral surface of the cylinder of radius R0 and
height h. Numerically, v 0 = 16 m s−1 . The same approach for the radius R1 indicates an air
velocity value v 1 = 1.6 m s−1 . For any value of the radius r, r being the radial coordinate, the
92
C de Izarra and G de Izarra
mean velocity is vm = Q/2π hr, and a possible mean value U of the air velocity evaluated
over the total surface of the CD is given by
R1
R1
2
Q
2
2π r dr
vm 2π r dr =
U π R1 − R0 =
2π
rh
R0
R0
that reduces to
Q
U=
.
π h(R1 + R0 )
Numerically, we obtain U = 2.6 m s−1 .
3. Theoretical approach
3.1. From the Navier–Stokes equation to the Stokes equation
Let us consider the Navier–Stokes equation for an incompressible fluid of specific mass ρ and
dynamical viscosity μ. We have, neglecting the external forces such as the weight,
∂ v
v = −∇P
+ μv .
ρ
+ ρ(v · ∇)
∂t
The reduced Reynolds number R ∗ defined in [1]
In the case of a permanent flow, ∂ v /∂t = 0.
gives the ratio of inertia to viscous forces, in the case of a system having two very different
typical lengths R1 and h, with R1 h:
v
ρ(v · ∇)
inertia force
=
.
R∗ =
viscous force
μv
The inertia force can be evaluated by considering the typical length R1, while the viscous force
is defined by considering the typical length h; using the averaged velocity U, we have, calling
ν = μ/ρ the kinematic viscosity,
ρU 2 /R1
U R1 h 2
∗
R =
=
.
μU/ h2
ν
R1
For air at ambient temperature and pressure, ν = 1.5 10−5 m2 s−1 , and the numerical value of
the reduced Reynolds number R ∗ is 0.72, tending to zero as h → 0. We can consider only
the viscous and the pressure terms in the Navier–Stokes equation, which reduces to the Stokes
equation, at the basis of lubrication theory [3].
3.2. Pressure distribution
The starting point is the Stokes equation
+ μv = 0
−∇P
(1)
and the continuity equation for an incompressible flow
· v = 0.
∇
(2)
where
Taking the cylindrical coordinates (r, ϕ, z), the velocity vector is v = vr er + vϕ eϕ + vz k,
is the orthonormal local basis of the cylindrical coordinate system. In our case, it is
(er , eϕ , k)
clear that the celerity components vϕ and vz are null. The radial component vr only depends
on the variables r and z. Using the definition of the Laplacian of a vector and the expression
of the gradient in cylindrical coordinates [4], equation (1) becomes
1 ∂P
∂P vr
∂P
(3)
er +
eϕ +
k = μ vr − 2 er ,
∂r
r ∂ϕ
∂z
r
Stokes equation in a toy CD hovercraft
with
vr =
1 ∂
r ∂r
r
93
∂vr
∂r
+
∂ 2 vr
∂z2
⇒
vr =
∂ 2 vr 1 ∂vr ∂ 2 vr
+
+
.
∂r 2
r ∂r
∂z2
(4)
From equation (2), we have
1 ∂ (rvr )
∂vr
vr
=0 ⇒
=− .
r ∂r
∂r
r
The operator ∂/∂r applied to equation (5) gives
∂ 2 vr
2 ∂vr
2vr
∂ 2 vr
=
−
=+ 2 .
⇒
∂r 2
r ∂r
∂r 2
r
Using the result given by equation (6), vr (equation (4)) becomes
vr ∂ 2 vr
+
.
r2
∂z2
Finally, equation (3) is reduced to
vr =
(5)
(6)
(7)
1 ∂P
∂P ∂P
∂ 2 vr
er +
eϕ +
(8)
k = μ 2 er .
∂r
r ∂ϕ
∂z
∂z
A quick analysis of equation (8) indicates that the pressure does not depend on the angle ϕ,
and is constant over the direction z, i.e. on the thickness of the air film. Integration of the
radial component is straightforward:
vr =
1 ∂P 2
z + Az + B
2μ ∂r
(9)
with the integration constants A and B found with the limit conditions vr (z = 0) = 0 and
vr (z = h) = 0. We obtain the parabolic shape
vr =
1 ∂P 2
(z − hz).
2μ ∂r
(10)
3.3. Lifting force
Knowing the velocity vr (z) from equation (10), it is possible to compute the volumetric air
flow Q:
Q=
vr · dS,
with dS the elementary lateral surface of a cylinder of radius r and height h. The calculations
give
π r ∂P 3
h.
6μ ∂r
Integration of the pressure gradient
Q=−
(11)
6μQ 1
∂P
=−
,
(12)
∂r
π h3 r
taking into account that the pressure P = PA for r = R1 , leads to the pressure distribution
P (r) plotted in figure 4:
6μQ
R1
.
(13)
P (r) = PA +
ln
3
πh
r
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C de Izarra and G de Izarra
1
0
Figure 4. Plot of the pressure in the air film under the toy CD hovercraft.
The lifting force F is obtained by integrating the overpressure δP (r) = P (r) − PA over
all the surface of the CD; let usunderline that the overpressure in the central area of radius R0
is constant, equal to 6μQ
ln RR10 . We have
πh3
F =
R1
R0
δP (r)2π r dr + π R02 δP (R0 )
or
F =
R1
R0
F =
6μQ
R1
2π r dr + π R02 δP (R0 ).
ln
3
πh
r
12μQ
h3
R1
r ln R1 dr −
R0
R1
R0
6μQ
R1
.
r ln r dr + π R02
ln
3
πh
R0
Integration of the second integral is made by parts; finally, we obtain
F =
3μQ 2
2
R
.
−
R
1
0
h3
(14)
The numerical evaluation of the force F with the experimental results proposed in
subsection 2.2 and μ = 1.85 × 10−5 kg m−1 s−1 is F = 0.35 N. This value must be compared
to the weight of the toy CD hovercraft (0.210 N); taking into account the experimental errors
(30% of relative error on h measurement), it is clear that the lifting force is of the same
magnitude as the weight of the system, and allows it to glide on a table.
95
y
Stokes equation in a toy CD hovercraft
0
1
2
3
4
Figure 5. Plot of the potential energy curve of the toy CD hovercraft.
3.4. Potential energy curve and stability
with C = 3μQ R12 − R02 , the
Writing the lifting force as a function of z F = C/z3 k,
) is, choosing V → 0 for z → ∞,
corresponding potential energy function V (F = −∇V
C
(15)
V = 2.
z
Taking into account the potential energy for the weight Vw = mgz, the total potential energy
of the system is
C
(16)
Vtot = 2 + mgz.
z
The plot of Vtot as a function of z is presented in figure 5; the minimum of potential energy
corresponds to a stable point of the system. Let us underline that vertical oscillations of the
toy CD hovercraft are possible around this stable point.
4. Study of the loaded hovercraft
In this experimental part, we have put different known weights of the toy CD hovercraft, and
we have measured the volumetric air flow Q. When the hovercraft glides on the table, the
lifting force is equal to the weight of the loaded hovercraft. Experimental results are presented
in figure 6. If we suppose that the mean velocity does not change radically for different loads
(F varies from 0.2 N to 0.8 N), then the volumetric air flow Q is proportional to h, and F is
proportional to Q/ h3 ; consequently, we can foresee the following law:
C
F = 2.
Q
96
C de Izarra and G de Izarra
1
0.8
-8
2
0.6
0.4
0.2
-4
-4
-4
-4
-4
3
Figure 6. Plot of the lifting force value against the volume air flow.
The best fit given in figure 6 is in good agreement with experimental results, considering the
experimental error mainly given by the measurements of the volume air flow.
5. Test of a bigger hovercraft
In this section, we present experimental tests obtained on a larger hovercraft, for which we
give the construction procedure.
5.1. Construction
According to [5, 6], we have built a larger toy hovercraft with a leaf blower. The basis of the
larger hovercraft is a disc of plywood (thickness 1 cm) of radius R1 = 0.6 m, and a circular
plastic sheet (chose a sheet of heavy plastic) of radius larger than the plywood disc (radius of
about 0.8 m) (figures 7(a) and (b)).
Half way between the centre of the disk and the edge, a hole is made in the plywood that
exactly fits the end of the leaf blower. The plywood disk is laid on the centre of the large
plastic sheet. The edges of the sheet are folded up over the plywood, and fixed to the top of the
plywood disc using a staple gun. Adhesive tape is used to tape the edge of the plastic down to
make it look good (figures 7(b) and (c)).
On the bottom side of the hovercraft, the plastic sheet must be fixed to the centre of the
plywood disc by using a rigid plastic disc (radius of about 5 cm) easily found on a coffee can
lid. The rigid plastic disc is fixed on the plywood disc using a staple gun (figure 7(d)).
Stokes equation in a toy CD hovercraft
Plywood disc
97
Plastic sheet
Plywood disc
R1
(a)
Top view
(b)
(c)
Rigid plastic disc
vent-holes
R0 = 15cm
Plastic sheet
(d)
Bottom view
Leaf blower
(f)
(e)
Figure 7. Larger hovercraft construction.
Figure 8. Larger hovercraft in operation.
With a razor knife, six vent-holes (radius of 4 cm) are cut in the plastic sheet (figure 7(e)).
They are placed on a circle of radius R0 = 0.15 m. The experiment indicates that it is necessary
to reinforce the thin necks of plastic between the holes using adhesive tape.
Flip the hovercraft over so that the plastic sheet is on the bottom and place it on a smooth
floor. Put the end of the leaf blower into the hole and turn it on. The plastic on the bottom
should inflate; the hovercraft lifts up slightly and starts gliding around (figure 8).
98
C de Izarra and G de Izarra
10 4
8000
6000
4000
2000
0
0
1
2
3
4
5
Figure 9. Plot of the lifting force value F (expression (14)) against the air film thickness h for the
larger hovercraft.
5.2. Results
The theoretical expressions found for the small CD hovercraft (lifting force) are valid for the
larger hovercraft. The maximum volumetric air flow indicated by the leaf blower constructor
is Q = 0.21 m3 s−1 , and the radii defined in the previous sections are R1 = 0.6 m and R0 =
0.15 m. The value of the thickness of the air film h between the ground surface and the plastic
sheet is difficult to evaluate with accuracy, and is located in the range [1 mm, 2mm]. However,
assuming that Q is constant, the plot of the theoretical value of the lifting force F (expression
(14)) as a function of h (figure 9) indicates rather high values of the lifting force F. It was
possible for us to lift four persons (mass ≈ 350 kg) on a linoleum surface, corresponding to a
lifting force of 3433 N.
6. Conclusion
This experimental study of the toy CD hovercraft is rich enough to illustrate the analysis
of the Navier–Stokes equation as a function of the Reynolds number value, in the case of
incompressible flows. It is used during the fluid mechanics course, and can lead to completely
analytical problems, difficult to find in the topics of fluid mechanics.
Acknowledgment
The authors wish to thank the anonymous reviewers for their constructive comments.
Stokes equation in a toy CD hovercraft
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